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QUARTERLY EXAMINATIONS
X Class Mathematics (English Medium)
MODEL PAPER
1
Time: 2  Hours Parts: A & B Max. Marks: 50
2
Time: 2 Hours Part - A Marks: 35

SECTION - I
Note: 1) Answer any 5 of the following questions choosing at least 2 from each group
A & B.
2) Each question carries 2 Marks (5 × 2 = 10)

et
Group - A

.)n
1. Define conjunction and write truth table.
2. If A, B are two sets. Prove that A' - B' = B - A

3.
h(a x+3 3x + 3
If f : R − {3} → R is defined by f(x) =  Show that f  = x (x ≠ 1)

tib
x-3 x-1
4. Find the value of k so that x3 - 3x2 + 4x + k is exactly divisible by (x - 2)

pra Group - B

√3
5.

du
In an equilateral triangle with side 'a', prove that the area of the triangle is a2
4

na
6. The mean of 10 observations is 16.3. By an error, one observation is registered

ee)
as 32 instead of 23. Find the correct mean.

7. ( )( .) (
w
1 3
If 0 1
2
–1 =
p
-1 Find p.
8.

w w
Write the merits of the Arithmetic mean.
SECTION - II
Note: 1) Answer any 4 of the following questions
2) Each question carries 1 Mark (4 × 1 = 4)
9. n(A∪B) = 51, n(A) = 20, n(B) = 44 then find n(A∩B).
10. Define constant function
 
11. Find the sum and product of the roots of the equation √ 3x2 + 9x + 6√ 3 = 0
12. Determine 'k' so that k + 2, 4k - 6 and 3k - 2 are the three consecutive terms
of an A.P.


13. The mean and median of a unimodal grouped data are 72.5 and 73.9 respec-
tively. Find the mode of the data.
1 4
14. If A = 2 1
( ) find A2
.
SECTION - III
Note: 1) Answer any 4 of the following questions choosing at least two from each
group A & B. Each question carries 4 Marks (4× 4 = 16)
Group - A
15. For any three sets A, B, C prove that A∪ (B∩C) = (A∪B) ∩ (A∪C).
16. Let f, g, h be functions defined by f(x) = x + 2, g(x) = 3x - 1 and h(x) = 2x, show
that ho (gof) = (hog) of.

17. ( 5 9
Find the independent term of x in the expansion of 3x −  ) et
a. 1 1 1
x2
n
If (b+c) (c+a) and (a+b) are in H.P. show  ,  ,  will also be in H.P.

h
18.
a2 b2 c2

tib
Group - B

ra
19. Define 'Thales theorem' and prove it.
20.
p
Find the median of the marks scored by 50 students in a 50 Marks test

u
d
Marks 1 - 10 11 - 20 21 - 30 31 - 40 41 - 50

na
No. of students 3 12 16 14 5

ee
21. Using matrix inversion method, solve the linear equations

.
2x + 5y = 11, 4x - 3y = 9

w ) ( )
w(
-2 1 2 0
22. Given A = 3 -1 , B = 5 -3 prove that (AB)−1 = B−1 A−1

w SECTION - IV
Note: Answer any one of the following. Each question carries 5 marks. (1 $ 5 = 5)
23. Construct a triangle ABC in which BC = 4 Cm, A = 50°
− and altitude
through A = 3 Cm.
24. Using graph of y = x2, solve the equation x2 − 4x + 3 = 0


PART - B
Max. Marks: 15 Time: 30 Minutes
Note: Pick out the correct answer from the choice. Each question carries 1/2 mark.
1. 'For some' symbol of quantifier ( )
⊃
A) B) ∈ C) ∨
 D) ∃
2. (A∪ B)' = ( )
A) A' ∪ B' B) A' ∩ B' C) A∪ B D) A∩B
3. For the following which one is bijection? ( )
A f B A f B A f B A f B
a x a x a x a > x

et
> >
> > >
A) b y B) b > y C) b > y D) b > y
>

.n
c z c > z c > z c z

ha
If f(x) = x2 - 3x + 2, f(-2) =

tib
4. ( )
A) 12 B) 4 C) -12 D) 0

ra
5. Discriminent of 2x2 - 7x + 3 = 0 is ( )
A) 20

up B) 24 C) 25 D) 26

d
6. 'n' arithmetic means are there in between a and b, Then d = ( )

na
a-b b−a a+b b-a
A)  B)  C)  D) 
n+1 n+1 n-1 n-1
7.

.ee
Angle in a semi circle is ( )

w
A) 0° B) 90° C) 180° D) 360°

w
8. Mean of 12, 15, x, 19, 25, 44 is 25 then x = ( )

w 
A) 20 B) 25 C) 30 D) 35

9. 
If
2 −4
d 5
= 14 then d = ( )

A) −1 B) 1 C) 2 D) 4
10. ∆ ABC ∼ ∆PQR, A = 50°, B = 60° then R = ( )
A) 50° B) 60° C) 70° D) 80°
II. Fill in the Blanks
11. B and C are disjoint sets. (A - B) ∪ (A - C) = ------
12. 'I' is identity function, I-1 (4) = -----


13. p Λ ∼p is example of ---
14. K a, K b, K c (K ≠ 0) are in G.P. then a, b, c are in -----
x x x x
15. Median of  , x, , ,  (x > 0) is 8 then x = -----
5 4 2 3
16. Harmonic Mean of x & y is -----
17. Value of x2 − x − 2 < 0 is in between -----, -----

( )
18. If P = 1 0 then P −1 = -----
0 1
19. Angles in the same segment of a circle are -----
20. If AT = A , A is called ----- Matrix.

et
III. Match the following

.n
(i) Group A Group B
⊃
B then A ∩ B =

a
21. A ( ) A) B
22.
h
In a G.P. a = 2, S∞ = 6 then r = ( ) B) A

tib
23. A∪A' = ( ) C) ∅

ra
24. If f(x) = x + 2, g(x) = x, fog(x) = ( ) D) µ

p
25. Quadrants of y = mx2 (m > 0) ( ) E) x + 2

du F) I, IV
2

na
G) 
3
H) I, II
(ii)

.ee
Group A Group B

w
26. A.M. of a + 2, a, a − 2 ( ) I) 60

28.
w )
27. 70, 60, 70, 80, 60, 100, 60 mode

w )(
(1 0
0 1
0
1
1
0
=
(

(
)

)
J) 90°

K) a

29. In a ∆ ABC BC2 + AB2 = AC2 then B= ( ) L)
( )
1 0
0 0

30. If A =
( ) ( )
2
6
8
5
,B=
2 x
6 5
A = B then x = ( ) M) I
N) 70
O) 8

P)
( )
0 1
1 0

Answers
SECTION - I
Group - A
1. Define Conjunction and write truth table.
Sol: If two or more statements are combined with connective 'and' then the Compound
Statement is said to be conjunction. This is denoted by p Λ q.
Ex: 2 is even and 2 prime number
Definition of Truth table. If p & q both are true then p Λ q is true.
Table: p q pΛq

et
T T T

.n
T F F
F T F

ha F F F

tib
2. If A, B are two sets. prove that A' − B' = B − A

ra
Sol: To Prove

p
B−A
⊃
(i) A' - B'

du
(ii) B − A ⊃ A' - B'

na
(i) x ∈ A' - B'
⇒ x ∈ A' and x ∉ B' ⇒ x ∉ A and x ∈ B

ee
⇒x∈B−A

.
∴ A' - B'
⊃
B − A ------ (1)

ww
(ii) x ∈ B − A
⇒ x ∈ B and x ∉ A ⇒ x ∉ B' and x ∈ A'

w
⇒ x ∈ A' - B'
∴B−A ⊃ A' - B' ------- (2)
from (1) & (2) A' - B' = B − A

3.
x+3 3x +3
( )
If f : R-{3} ->R is defined by f(x) =  show that f  = x (x ≠ 1)
x−3 x −1
3x+3
+3
Sol: LHS= f  ( )
3x + 3
x−1
= x−1
3x + 3
 −3
x−1


3x + 3 + 3 (x − 1) / x − 1
= 
3x + 3 − 3(x − 1) / x − 1
3x + 3 + 3x − 3
= 
3x + 3 − 3x + 3
6x
=  = x RHS
6
3x + 3
( )
∴f  = x
x−1
4. Find the value of k so that x3 − 3x2 + 4x + k is exactly divisible by (x − 2).
Sol: f(x) = x3 − 3x2 + 4x + k

et
f(x) is exactly divisible by x − 2
∴ f(2) = 0

.n
f(2) = 23 – 3(2)2 + 4(2) + k = 0

a
h
8 – 12 + 8 + k = 0

tib
16 − 12 + k = 0

ra
4+k=0

p
∴ k = −4

du Group - B

√3 2

na
5. In an equilateral triangle with side 'a', prove that the area of the triangle is a .
4

ee
Sol: side an equilateral triangle = a

.
A
In ∆ ABC AB2 = AD2 + BD2

ww)
( a 2
a2 = x2 +  a
x
a

w
2
a2 4a2−a2 3a2
x 2 = a2 −  =  = 
4 4 4 B a/ D a/ C

2 2

∴Height: x = √ 3a2 √ 3 a
=
4 2
1
∴ Area of a equilateral triangle 
=2 × bh

1 √3 a
=×a×
2 2



√ 3a2
=  Sq. units.
4
6. The mean of 10 observations is 16.3 By an error, one observation is registered as
32 instead of 23. Find the correct mean.
Sol: The mean of 10 observations = 16.3
Sum of 10 observations = 16.3 × 10 = 163
Registered as 32 instead of 23
163 – 32 + 23 = 154
154
∴ Correct men  15.4
10
= =

et
1 3 2 p
=

.n
7. 0 1
( )( ) ( ) −1 −1 then p = ?

a
1 3 2 p
=

h
Sol: 0 1 −1 −1

tib
( )( ) ( )
1 × 2 + 3(−1)
[ ][]
0 × 2 + 1(−1) =
p
−1
2−3
[ ][]
0−1 =
pra p
−1
−1
[ ][] du p

na
−1 = −1

ee
⇒ p = −1

.
8. Write the merits of Arithmetic mean.

ww
Sol: i) It is uniquely defined. i.e., it has one and only one value
ii) It is based on all observations

w iii) it is easily understood
iv) it is easy to compute
SECTION − II
9. If n(A∪B) = 51, n(A) = 20, n(B) = 44 then find n(A∩B) = ?
Sol: n(A∪B) = n(A) + n(B) - n(A∩B)
51 = 20 + 44 - n(A∩B)
51 = 64 - n(A∩B)
∴ n(A∩B) = 64 - 51 = 13
10. Define Constant function.

Sol: A function f : A → B is a constant function if there is an element c ∈ B such that
f(x) = C for all x ∈ A
Ex: f = {(x, 5) / x ∈ N}
 
11. Find the sum and product of the roots of the √ 3x2 + 9x + 6√ 3 = 0
 
Sol: a = √ 3, b = 9, c = 6√ 3
 
−b −9 √ 3 −9√ 3 
The sum of roots  =  ×  =  = −3√ 3
 
a √3 √3
=
3

c 6√ 3
The product of roots = =6

a √3
=
12. Determine k so that k + 2, 4k − 6 and 3k − 2 are the three consecutive terms of

et
an A.P.

.n
Sol: 4k − 6 − (k + 2) = (3k − 2) − (4k − 6)

a
⇒ 4k − 6 − k - 2 = 3k − 2 − 4k + 6

h
tib
⇒ 3k − 8 = k + 4
⇒ 3k + k = 4 + 8
⇒ 4k = 12

pra
u
12
∴k==3
4
d
na
13. The mean and median of a unimodal grouped data are 72.5 and 73.9 respective-

ee
ly. Find the mode of the data.

.
Sol: Mode = 3 × median - 2 × mean

ww = 3(73.9) -2 (72.5)
= 221.7 -145

w( )
14. If A =
= 76.7
1 4
2 1 Find
A2.

1 4 1 4
Sol: A2 = A × A = 2 1
( )( ) 2 1
1×1+4×2 1×4+4×1
= ( )
2×1+1×2 2×4+1×1
1+8 4+4
= 2+2 8+1
( )

9 8
= 4 9
( ) SECTION - III
Group - A
15. For any 3 sets A, B, C prove that A∪(B∩C) = (A∪B)∩(A∪C)
⊃
Sol: We have to prove that: (i) A∪(B∩C) (A∪B)∩(A∪C)
⊃
(ii) (A∪B)∩(A∪C) A∪(B∩C).
(i) x ∈ A∪(B∩C)
⇒ x ∈ A or x ∈ (B∩C)
⇒ x ∈ A or (x ∈ B and x ∈ C)

et
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)

.n
⇒ x ∈ (A∪B) ∩ (A∪C)
∴ A∪(B∩C)
a (A∪B) ∩ (A∪C) ------ (1)
⊃

h
tib
(ii) x ∈ (A∪B) ∩ (A∪C)
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)

pra
⇒ x ∈ A or (x ∈ B and x ∈ C)
⇒ x ∈ A or x ∈ (B ∩ C)

du
⇒ x ∈ A∪(B∩C)

na
∴ (A∪B) ∩ (A∪C) A ∪ (B∩C) ------- (2)
⊃

ee
From (1) & (2) A ∪ (B∩C) = (A∪B) ∩ (A∪C)
16.

w.
Let f, g, h functions be defined by f(x) = x + 2, g(x) = 3x − 1 & h(x) = 2x. Then
show that ho(gof) = (hog)of

ww
Sol: (i) ho(gof)(x)
gof(x) = g[f(x)]
= g[x + 2]
= 3(x + 2) − 1
= 3x + 6 − 1
= 3x + 5
ho(gof)(x) = h[gof(x)]
= h[3x + 5]
= 2(3x + 5)
= 6x+ 10 -------- (1)


(ii) (hog)of(x)
hog(x) = h[g(x)]
= h[3x − 1]
= 2(3x − 1)
= 6x − 2
(hog)of(x) = hog[f(x)]
= hog(x + 2)
= 6(x + 2) − 2
= 6x + 12 − 2

et
= 6x + 10 -------- (2)
From (1) & (2) ho(gof) = (hog)of

( ) .n
17. Find the Independent term in the expansion of 3x− 
5 9

−5
ha x2

tib
Sol: a = 3x, b = , n = 9, r = r
x2

ra
−
Tr+1 = nc xn−r yr
r

( ) up
= 9Cr (3x)9−r 
−5 r

d
x2

na
= 9Cr (3)9−r (x)9−r (−5)r. x−2r

ee
= 9Cr (3)9−r (−5)r. x9−r−2r

.
= 9Cr (3)9−r (−5)r. x9−3r

w
We have to find independent term

ww∴ Power of x = 0
9 − 3r = 0
9
3r = 9 ⇒ r =  = 3
3
∴T3+1 = 9C3 (3x)9−3 
x2
( )
−5 3

(−5)3
T4 = 9C3 (3)6. x6. 
x6
T4 = 9C3 (3)6 (−5)3


1 1 1
18. If (b + c), (c + a) & (a + b) are in H.P. then show that , ,  will also be
a2 b2 c2
in H.P.
Sol: (b + c), (c + a), (a + b) --- are in H.P
1 1 1
 ,  ,  --- are in A.P.
b+c c+a a+b
1 1 1 1
∴− =  − 
c+a b+c a+b c+a
b+c−c−a
/ / c+a−a−b
/ /
 = 
/
(c + a) (b + c) /
(a + b) (c + a)
b−a c−b

et
 = b+a
c+b 

.n
by cross multiplication

ha
(b − a) (b + a) = (c − b) (c + b)
b2 − a2 = c2 − b2

tib
∴ a2, b2, c2 are in A.P.

ra
1 1 1
∴ , ,  are in H.P.
a2 b2 c2

up Group − B
19.
d
Define 'Thales theorem' and prove it.

na
Sol: In a triangle, a line drawn parallel to one side, will divide the other two sides in

ee
the same ratio.

.
Given: ∆ABC, in Which DE//BC and DE intersects AB in D and AC in E

w
AD AE
To prove:  = 

w
DB EC A

w
Construction: Join BE, CD and draw EF ⊥ BA
Proof: In ∆ADE, ∆BDE
1
∆ADE  × AD.EF AD D
F
E
2
 =  =  ------- (1)
∆BDE 1
 × BD.EF DB
2 >
∆AED AE B C
similarly:  =  ------- (2)
∆ECD EC
But ∆BDE = ∆ECD
(Triangles on the same base DE, and between the same parallel lines DE & BC)


AD AE
From (1) & (2) We get  = 
DB EC
20. Find the median of the marks scored by 50 students in a 50 marks test.
Marks 1 - 10 11 - 20 21 - 30 31 - 40 41 - 50
No. of 3 12 16 14 5
Students
Sol:
CI f Cum.Frequency
N 50
1 - 10 3 3  =  = 25
2 2
11 - 20 12 15F

et
21 - 30 16f 31
L = 20.5 Median class
31 - 40 14 45

a.n 41 - 50 5 50

h
N = 50

( ) ra tib
N
-F
2 ×C
Median = L+ 
f

( ) dup
= 20.5 +
25 - 15
 × 10

na
16
100

ee
= 20.5 + 
16

w.
= 20.5 + 6.25
= 26.75
21.

ww
Using Matrix inversion method, solve the linear equation 2x + 5y = 11,
4x-3y=9
Sol: 4x - 3y = 9
2x + 5y = 11

A= ( ) () ()
2 5
4 −3
X=
x
y
B=
11
9
AX = B ⇒ X = A−1 B
A = 2(−3) − 5 × 4 = −6 − 20 = −26 ≠ 0
∴ A−1 exists.


d −b
A−1 = 
1
ad − bc ( ) −c a
−3 −5
= 
1
2(−3)−5 × 4 [ ] −4 2

[ ]
3 5
 
−3 −5
= 
1
−6 −20 [ ]
−4 2
=
26
4
26
26
−2

26
X = A−1 B

( )( )
3 5
 
26 26 11
=
4 −2
  9

et
26 26
 + 45
33 78

() ( ) ( ) ()  

.n
x 26 26 26 3
X= = = =

a
y 44 − 18 26 1
  

h
26 26 26

tib
⇒ x = 3, y = 1

ra
−2 1 2 0
( ) ( )
22. Given = 3 −1 , B = 5 −3 then Prove that (AB)−1 = B−1A−1.

( up)
(−2)(2) + (1)(5) (−2)(0) + (1)(−3)
Sol: AB =
d
na( )
(3)(2) + (−1)(5) (3)(0) + (−1)(−3)

( ee )
=
−4 + 5 0 −3
=
1 3

w . ( )
(AB)−1 = 
6−5
1
0+3 −1
3 3
3

ww (1)(3) − (−3)(1) −1 3

( ) [ ]
3 3
1 3 3  
=  = 6 6
3+3 −1 1 −1 1

6 6

( )( )
3

( )
B=
2
5 −3
0
⇒ B−1 = 
1 −3 0
(2)(−3) − (0)(5) −5 2
= 6
5

6
0
2
−
6

( )
A=
−2 1
3 −1 ( )⇒ A−1 = 
1
(−2)(−1) − (1)(3)
−1
−3
−1
−2


1
= 
2 −3 ( −1
−3
−1
−2 ) ( )
=
1
3
1
2

( )( )
3
 0 1 1
B−1 A−1 = 6
 −
5 2 3 2
6 6

( )( )
3 3 3 3
 +0  +0  
= 6 6 = 6 6
5 −
6 5 − 
4 −  
1 1
 
6 6 6 6 6 6
∴ (AB)−1 = B−1 A−1..

et
SECTION - IV
23. Construct a triangle ABC in which BC = 4 cm, A = 50°. and altitude through
−

.n
A = 3 cm.

ha
tib
pra
du
na
.ee
ww
w
Construction:
(1) Draw a line segment BC = 4 cm and make an angle PBC = 50° with the help of
a protractor.
(2) Draw perpendicular bisector RQ of BC. Draw perpendicular EB to BP. Let RQ
and EB intersect in a point say 'O'. Let M be mid point of BC.
(3) Taking 'O' as centre and OB as radius draw a circle
(4) Take a point L on RQ such that the line segment ML = 3 cm
(5) Draw AA' // BC through L intersecting the circle in two points say A and A'. Join
AB, AC and A'B, A'C.
Either of the triangle ABC, A'BC will be the required triangle.


24. Using graph of y = x2, solve the equation x2 − 4x + 3 = 0.

et
Sol. x2 − 4x + 3 = 0; x2 = 4x − 3. y = x2 is parabola and y = 4x − 3 is straight line

.n
y = x2
x 0

ha 1 2 3 −1 −2 −3

ib
y 0 1 4 9 1 4 9

t
ra
y = 4x − 3

p
x 0 1 2 3 −1

du y −3 1 5 9 −7

a
.een
ww
w


PART - B

1) D
.n
2) B et Answers
3) B 4) A
5) C

ha
6) B 7) B 8) D

tib
9) B 10) C 11) A 12) 4

ra
2xy
13) Contradiction 14) Arithmetic progression 15) 24 16) 

( )dup
x+y
1 0
17) -1, 2 18) 19) Equal 20) Symmetric matrix
0 1

na
21) A 22) G 23) D 24) E

.e
25) H
29) J
e 26) K
30) O
27) I 28) P

ww Writer: P.Venugopal

w
S.A.(Maths) Govt. TW AHS (Girls),
Jaggannapet, Warangal District


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